Problem: You have found the following ages (in years) of all 6 lions at your local zoo: $ 6,\enspace 2,\enspace 10,\enspace 4,\enspace 3,\enspace 10$ What is the average age of the lions at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{6 + 2 + 10 + 4 + 3 + 10}{{6}} = {5.8\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $6$ years $0.2$ years $0.04$ years $^2$ $2$ years $-3.8$ years $14.44$ years $^2$ $10$ years $4.2$ years $17.64$ years $^2$ $4$ years $-1.8$ years $3.24$ years $^2$ $3$ years $-2.8$ years $7.84$ years $^2$ $10$ years $4.2$ years $17.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.04} + {14.44} + {17.64} + {3.24} + {7.84} + {17.64}} {{6}} $ $ {\sigma^2} = \dfrac{{60.84}}{{6}} = {10.14\text{ years}^2} $ The average lion at the zoo is 5.8 years old. The population variance is 10.14 years $^2$.